Question: For each integer $n$ greater than 1, let $F(n)$ be the number of solutions of the equation $\sin x = \sin nx$ on the interval $[0, \pi]$.  What is $\sum_{n=2}^{2007} F(n)$?
Answer: Note that $F(n)$ is the number of points at which the graphs of $y=\sin x$ and $y=\sin nx$ intersect on $[0,\pi]$.  For each $n$, $\sin nx \geq 0$ on each interval $\left[ \frac{(2k-2) \pi}{n}, \frac{(2k-1) \pi}{n} \right]$ where $k $ is a positive integer and $2k-1 \leq n$.  The number of such intervals is $\frac{n}{2}$ if $n$ is even and $\frac{n + 1}{2}$ if $n$ is odd.

The graphs intersect twice on each interval unless $\sin x = 1 = \sin nx$ at some point in the interval, in which case the graphs intersect once. This last equation is satisfied if and only if $n \equiv 1\pmod 4$ and the interval contains $\frac{\pi}{2}$.  If $n$ is even, this count does not include the point of intersection at $(\pi,0)$.

Therefore $F(n)= 2 \cdot \frac{n}{2} + 1=n+1$ if $n$ is even, $F(n)=\frac{2(n+1)}{2}=n+1$ if $n \equiv 3\pmod 4$, and $F(n)=n$ if $n \equiv 1\pmod 4$.  Hence,
\[\sum_{n=2}^{2007} F(n)=\left(\sum_{n=2}^{2007} (n+1)\right) - \left\lfloor \frac{2007-1}{4}\right\rfloor = \frac{(2006)(3+2008)}{2}-501 = \boxed{2{,}016{,}532}.\]